Problem 62 - Applied Calculus

Question

Product Recognition A company begins advertising a new product and finds that after \(t\) weeks the product is gaining customer recognition at the rate of \(t^{2} \ln t\) thousand customers per week (for \(t \geq 1\) ). Find the total gain in recognition from the end of week 1 to the end of week \(6 .\)

Step-by-Step Solution

Answer

Approximately 72\(\ln 6\) - \(\frac{647}{9}\) thousand customers.

Step 1: Understand the Problem

We need to find the total gain in customer recognition from the end of week 1 to the end of week 6. The rate of gaining recognition is given by the function \( R(t) = t^2 \ln t \) (in thousands of customers per week).

Step 2: Set up the Integral

To find the total gain in recognition, we need to integrate the rate function from week 1 to week 6. This means setting up the integral: \[\int_{1}^{6} t^2 \ln t \, dt \]

Step 3: Apply Integration by Parts

To solve the integral \( \int t^2 \ln t \, dt \), we'll use integration by parts, where \( u = \ln t \) and \( dv = t^2 \, dt \). Then, \( du = \frac{1}{t} \, dt \) and \( v = \frac{t^3}{3} \). Apply integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]

Step 4: Evaluate Parts

Substitute the parts: \[ uv = \ln t \cdot \frac{t^3}{3} = \frac{t^3 \ln t}{3} \] \[ \int v \, du = \int \frac{t^3}{3} \cdot \frac{1}{t} \, dt = \int \frac{t^2}{3} \, dt = \frac{t^3}{9} + C \]

Step 5: Construct the Antiderivative Function

Now combine these parts into the antiderivative: \[ \int t^2 \ln t \, dt = \frac{t^3 \ln t}{3} - \frac{t^3}{9} + C \]

Step 6: Evaluate the Definite Integral

Evaluate the definite integral from 1 to 6 for the antiderivative: \[ \left[ \frac{t^3 \ln t}{3} - \frac{t^3}{9} \right]_{1}^{6} \] First, substitute \( t = 6 \) and then \( t = 1 \), and subtract the values.

Step 7: Calculate the Definite Integral

Substituting \( t = 6 \):\[ \frac{6^3 \ln 6}{3} - \frac{6^3}{9} = 72 \ln 6 - 72 \]Substituting \( t = 1 \):\[ \frac{1^3 \ln 1}{3} - \frac{1^3}{9} = 0 - \frac{1}{9} = -\frac{1}{9} \]Resulting in:\[ \left( 72 \ln 6 - 72 \right) - \left( -\frac{1}{9} \right) = 72 \ln 6 - 72 + \frac{1}{9} \]

Step 8: Conclusion

The calculated recognition gain is approximately \( 72 \ln 6 - \frac{647}{9} \) thousand customers, which is the number achieved through calculation.

Key Concepts

Integration by Parts

Integration by parts is a technique for solving integrals, especially useful when dealing with products of functions. The formula used is:

• \( \int u \, dv = uv - \int v \, du \)

This concept originates from the product rule in differentiation and aims to simplify complex integrals. When using integration by parts:

• Choose two parts of the integrand, \(u\) and \(dv\).
• Differentiate \(u\) to get \(du\), and integrate \(dv\) to find \(v\).

Apply it step-by-step:

• Calculate \(uv\).
• Then solve \(\int v \, du\).
• Combine results for the solution.

In the context of our exercise, \(u\) is \(\ln t\) and \(dv\) is \(t^2 \, dt\). Consequently, \(du\) is \(\frac{1}{t} \, dt\), and \(v\) becomes \( \frac{t^3}{3} \). This method allowed us to tackle the function \(t^2 \ln t\) efficiently.

Definite Integral

A definite integral calculates the area under a curve between two boundaries or limits. It's expressed through:

• \( \int_{a}^{b} f(x) \, dx \)

The value obtained with a definite integral reflects the total accumulation of the function over an interval between \(a\) and \(b\). This is key for understanding quantities that accumulate over time, like distance, area, or, in our exercise, customer recognition. When evaluating:

• First, find the antiderivative of the function \(f(x)\).
• Next, substitute the values of the upper limit \(b\) and the lower limit \(a\) into the antiderivative.
• Subtract these to find the total area or, in our case, the total gain in customer recognition.

For the given problem, it was necessary to integrate from \(1\) to \(6\), applying what we calculated from the integration by parts to find the overall increase in customer recognition over these weeks.

Antiderivative

An antiderivative, or indefinite integral, is the reverse of differentiation. Instead of finding the slope, we focus on finding the original function whose derivative is given.To find an antiderivative:

• Identify basic function forms whose derivatives match parts of your integrand.
• Adjust coefficients if necessary.
• Look for common integral patterns, like powers or logarithms.

The process involves determining a function \(F(x)\) such that \(F'(x) = f(x)\). In our exercise, we used integration by parts to uncover an antiderivative for \(t^2 \ln t\). We combined sub-results to build a single expression: \( \frac{t^3 \ln t}{3} - \frac{t^3}{9} \), which served as our antiderivative for integration. Successfully finding this expression was crucial for proceeding with the final evaluation of the definite integral.